Question

The electron mobility in a conductor is \(32~\text{cm}^{2}/\text{Vs}\). What is the relaxation time of free electrons? (Given: \(e=1.6\times10^{-19}~C,\; m=9.1\times10^{-31}~kg\))

• A. \(1.82\times10^{-12}\;s\)
• B. \(2.73\times10^{-12}\;s\)
• C. \(1.82\times10^{-14}\;s\)
• D. \(2.73\times10^{-14}\;s\)

Hint:

For questions involving electron mobility and relaxation time, always remember the relation \[ \boxed{\mu=\frac{e\tau}{m}} \] and convert mobility from \(\text{cm}^{2}/\text{Vs}\) to \(\text{m}^{2}/\text{Vs}\) before substitution. Useful conversion: \[ 1~\text{cm}^{2}=10^{-4}~\text{m}^{2}. \]

Answer 1

The Correct Option is A

Electron mobility is related to relaxation time by $\mu = e\tau/m$, so $\tau = \mu m/e$. Substituting $\mu = 32 \times 10^{-4}$ m²/Vs, $m = 9.1 \times 10^{-31}$ kg, and $e = 1.6 \times 10^{-19}$ C gives $\tau \approx 1.82 \times 10^{-13}$ s.